Question

How do you solve `log_5(3x-1)=log_5 (2x^2)` and check the solutions?

Answer 1

`x = 1/2 and 1`

Explanation:

We start by using the property that if `color(magenta)(log_a b = log_a c`, then `color(magenta)(b = c)`.

Therefore:

`3x - 1 = 2x^2`

`0 = 2x^2 - 3x + 1`

`0 = 2x^2 - 2x - x + 1`

`0 = 2x(x- 1) - 1(x - 1)`

`0 = (2x- 1)(x - 1)`

`x = 1/2 and 1`

Checking in the original equation, you will find both solutions work. Note that our restrictions on the variable are `x > 0`, because the log function is undefined in the real number system whenever the value within the logarithm is equal to or smaller than `0`. We can automatically confirm both solutions are correct because neither contradict the restriction.

Hopefully this helps!